Integrand size = 25, antiderivative size = 254 \[ \int \frac {1+x+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{-x+2^{2/3} \sqrt [3]{x^2+x^4}}\right )}{4 \sqrt [3]{2}}-\frac {3 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{x^2+x^4}}\right )}{4 \sqrt [3]{2}}+\frac {3 \log \left (-2 x+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{4 \sqrt [3]{2}}-\frac {\log \left (2 x+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{4 \sqrt [3]{2}}+\frac {\log \left (-2 x^2+2^{2/3} x \sqrt [3]{x^2+x^4}-\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{8 \sqrt [3]{2}}-\frac {3 \log \left (2 x^2+2^{2/3} x \sqrt [3]{x^2+x^4}+\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{8 \sqrt [3]{2}} \]
-1/8*3^(1/2)*arctan(3^(1/2)*x/(-x+2^(2/3)*(x^4+x^2)^(1/3)))*2^(2/3)-3/8*3^ (1/2)*arctan(3^(1/2)*x/(x+2^(2/3)*(x^4+x^2)^(1/3)))*2^(2/3)+3/8*ln(-2*x+2^ (2/3)*(x^4+x^2)^(1/3))*2^(2/3)-1/8*ln(2*x+2^(2/3)*(x^4+x^2)^(1/3))*2^(2/3) +1/16*ln(-2*x^2+2^(2/3)*x*(x^4+x^2)^(1/3)-2^(1/3)*(x^4+x^2)^(2/3))*2^(2/3) -3/16*ln(2*x^2+2^(2/3)*x*(x^4+x^2)^(1/3)+2^(1/3)*(x^4+x^2)^(2/3))*2^(2/3)
Time = 1.02 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.04 \[ \int \frac {1+x+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=\frac {x^{2/3} \sqrt [3]{1+x^2} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}-2^{2/3} \sqrt [3]{1+x^2}}\right )-6 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}}\right )+6 \log \left (-2 \sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}\right )-2 \log \left (2 \sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}\right )+\log \left (-2 x^{2/3}+2^{2/3} \sqrt [3]{x} \sqrt [3]{1+x^2}-\sqrt [3]{2} \left (1+x^2\right )^{2/3}\right )-3 \log \left (2 x^{2/3}+2^{2/3} \sqrt [3]{x} \sqrt [3]{1+x^2}+\sqrt [3]{2} \left (1+x^2\right )^{2/3}\right )\right )}{8 \sqrt [3]{2} \sqrt [3]{x^2+x^4}} \]
(x^(2/3)*(1 + x^2)^(1/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*x^(1/3))/(x^(1/3) - 2^ (2/3)*(1 + x^2)^(1/3))] - 6*Sqrt[3]*ArcTan[(Sqrt[3]*x^(1/3))/(x^(1/3) + 2^ (2/3)*(1 + x^2)^(1/3))] + 6*Log[-2*x^(1/3) + 2^(2/3)*(1 + x^2)^(1/3)] - 2* Log[2*x^(1/3) + 2^(2/3)*(1 + x^2)^(1/3)] + Log[-2*x^(2/3) + 2^(2/3)*x^(1/3 )*(1 + x^2)^(1/3) - 2^(1/3)*(1 + x^2)^(2/3)] - 3*Log[2*x^(2/3) + 2^(2/3)*x ^(1/3)*(1 + x^2)^(1/3) + 2^(1/3)*(1 + x^2)^(2/3)]))/(8*2^(1/3)*(x^2 + x^4) ^(1/3))
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 1.14 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.63, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2467, 25, 2035, 7276, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x^2+x+1}{\left (x^2-1\right ) \sqrt [3]{x^4+x^2}} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {x^{2/3} \sqrt [3]{x^2+1} \int -\frac {x^2+x+1}{x^{2/3} \left (1-x^2\right ) \sqrt [3]{x^2+1}}dx}{\sqrt [3]{x^4+x^2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {x^{2/3} \sqrt [3]{x^2+1} \int \frac {x^2+x+1}{x^{2/3} \left (1-x^2\right ) \sqrt [3]{x^2+1}}dx}{\sqrt [3]{x^4+x^2}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 x^{2/3} \sqrt [3]{x^2+1} \int \frac {x^2+x+1}{\left (1-x^2\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}}{\sqrt [3]{x^4+x^2}}\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle -\frac {3 x^{2/3} \sqrt [3]{x^2+1} \int \left (\frac {x+2}{\left (1-x^2\right ) \sqrt [3]{x^2+1}}-\frac {1}{\sqrt [3]{x^2+1}}\right )d\sqrt [3]{x}}{\sqrt [3]{x^4+x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 x^{2/3} \sqrt [3]{x^2+1} \left (2 \sqrt [3]{x} \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{3},\frac {7}{6},x^2,-x^2\right )+\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}}{\sqrt {3}}\right )}{4 \sqrt [3]{2}}-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}}{\sqrt {3}}\right )}{4 \sqrt [3]{2} \sqrt {3}}+\frac {\sqrt {3} \arctan \left (\frac {\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{8 \sqrt [3]{2}}-\frac {\arctan \left (\frac {\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{8 \sqrt [3]{2} \sqrt {3}}-\sqrt [3]{x} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {7}{6},-x^2\right )+\frac {\log \left (\left (1-x^{2/3}\right )^2 \left (x^{2/3}+1\right )\right )}{24 \sqrt [3]{2}}+\frac {\log \left (\frac {2^{2/3} \left (x^{2/3}+1\right )^2}{\left (x^2+1\right )^{2/3}}-\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1\right )}{12 \sqrt [3]{2}}-\frac {\log \left (\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1\right )}{6 \sqrt [3]{2}}-\frac {\log \left (x^{2/3}-2^{2/3} \sqrt [3]{x^2+1}+1\right )}{8 \sqrt [3]{2}}\right )}{\sqrt [3]{x^4+x^2}}\) |
(-3*x^(2/3)*(1 + x^2)^(1/3)*(2*x^(1/3)*AppellF1[1/6, 1, 1/3, 7/6, x^2, -x^ 2] - ArcTan[(1 - (2*2^(1/3)*(1 + x^(2/3)))/(1 + x^2)^(1/3))/Sqrt[3]]/(4*2^ (1/3)*Sqrt[3]) + (Sqrt[3]*ArcTan[(1 - (2*2^(1/3)*(1 + x^(2/3)))/(1 + x^2)^ (1/3))/Sqrt[3]])/(4*2^(1/3)) - ArcTan[(1 + (2^(1/3)*(1 + x^(2/3)))/(1 + x^ 2)^(1/3))/Sqrt[3]]/(8*2^(1/3)*Sqrt[3]) + (Sqrt[3]*ArcTan[(1 + (2^(1/3)*(1 + x^(2/3)))/(1 + x^2)^(1/3))/Sqrt[3]])/(8*2^(1/3)) - x^(1/3)*Hypergeometri c2F1[1/6, 1/3, 7/6, -x^2] + Log[(1 - x^(2/3))^2*(1 + x^(2/3))]/(24*2^(1/3) ) + Log[1 + (2^(2/3)*(1 + x^(2/3))^2)/(1 + x^2)^(2/3) - (2^(1/3)*(1 + x^(2 /3)))/(1 + x^2)^(1/3)]/(12*2^(1/3)) - Log[1 + (2^(1/3)*(1 + x^(2/3)))/(1 + x^2)^(1/3)]/(6*2^(1/3)) - Log[1 + x^(2/3) - 2^(2/3)*(1 + x^2)^(1/3)]/(8*2 ^(1/3))))/(x^2 + x^4)^(1/3)
3.28.38.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 34.95 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.79
| method | result | size |
| pseudoelliptic | \(-\frac {2^{\frac {2}{3}} \left (\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (-2^{\frac {2}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}+x \right )}{3 x}\right )-3 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}+x \right )}{3 x}\right )-\frac {\ln \left (\frac {2^{\frac {2}{3}} x^{2}-2^{\frac {1}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}+\frac {3 \ln \left (\frac {2^{\frac {2}{3}} x^{2}+2^{\frac {1}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}+\ln \left (\frac {2^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{x}\right )-3 \ln \left (\frac {-2^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{x}\right )\right )}{8}\) | \(201\) |
| trager | \(\text {Expression too large to display}\) | \(8780\) |
-1/8*2^(2/3)*(3^(1/2)*arctan(1/3*3^(1/2)*(-2^(2/3)*(x^2*(x^2+1))^(1/3)+x)/ x)-3*3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)*(x^2*(x^2+1))^(1/3)+x)/x)-1/2*ln( (2^(2/3)*x^2-2^(1/3)*(x^2*(x^2+1))^(1/3)*x+(x^2*(x^2+1))^(2/3))/x^2)+3/2*l n((2^(2/3)*x^2+2^(1/3)*(x^2*(x^2+1))^(1/3)*x+(x^2*(x^2+1))^(2/3))/x^2)+ln( (2^(1/3)*x+(x^2*(x^2+1))^(1/3))/x)-3*ln((-2^(1/3)*x+(x^2*(x^2+1))^(1/3))/x ))
\[ \int \frac {1+x+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=\int { \frac {x^{2} + x + 1}{{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} - 1\right )}} \,d x } \]
\[ \int \frac {1+x+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=\int \frac {x^{2} + x + 1}{\sqrt [3]{x^{2} \left (x^{2} + 1\right )} \left (x - 1\right ) \left (x + 1\right )}\, dx \]
\[ \int \frac {1+x+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=\int { \frac {x^{2} + x + 1}{{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} - 1\right )}} \,d x } \]
\[ \int \frac {1+x+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=\int { \frac {x^{2} + x + 1}{{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} - 1\right )}} \,d x } \]
Timed out. \[ \int \frac {1+x+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=\int \frac {x^2+x+1}{{\left (x^4+x^2\right )}^{1/3}\,\left (x^2-1\right )} \,d x \]